0
$\begingroup$

I have been ask to price a European option with payoff $H(X_T,T) = X_T^2$ using the equivalent martingale measure (EMM).

For this I used the process:

\begin{equation} dX_t = r X_t dt + \sigma X_t d\tilde{B}_t \end{equation}

which is a geometric brownian motion in the EMM.

For the pricing we have:

\begin{equation} E[e^{-r(T-t)}f(X_T)|\mathfrak{F}_s] = e^{-r(T-t)}E\left[\left(X_t e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \omega \sqrt{T-t}}\right)^2|\mathfrak{F}_t\right] \end{equation}

where $\tilde{B}_{T-t}$ was replace by $\omega \sqrt{T-t}$ as $\tilde{B}_{T-t} \sim \mathcal{N}(0,T-t)$.

The condition of the expected value can be taken of as the process is a martingale under measure $Q$ and there fore the expected value can be computed, and this is where I get stuck. I have the following.

\begin{equation} \begin{aligned} e^{-r(T-t)}E\left[\left(X_t e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \omega \sqrt{T-t}}\right)^2|\mathfrak{F}_t\right] = e^{-r(T-t)}E\left[\left(X_t e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \omega \sqrt{T-t}}\right)^2\right] \\ = e^{-r(T-t)} \frac{1}{\sqrt{2\pi}}\int_{R} \left(X_t e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \omega \sqrt{T-t}}\right)^2 e^{-\frac{1}{2} \omega^2}d\omega \\ = e^{-r(T-t)} \frac{1}{\sqrt{2\pi}}\int_{R} X_t^2 e^{(2r - \sigma^2)(T-t) + 2\sigma \omega\sqrt{T-t}-\frac{1}{2}\omega^2}d\omega \\ = e^{r(T-t)}X_t^2 \frac{1}{\sqrt{2\pi}}\int_R e^{-\sigma^2(T-t) + 2\sigma\omega\sqrt{T-t} - \frac{1}{2}\omega^2} d\omega \end{aligned} \end{equation}

In a normal payoff like $(X_T - K)^+$ we would the $\omega$ for which the integral is positive and then factor everything in the exponent. But in this case I'm not sure how to continue as I'm not seeing an easy solution to the integral.

Improve this question
$\endgroup$
7
  • $\begingroup$ Is the payoff $(X_T^2-K)^+$ or $X_T^2$? $\endgroup$
    – ForeignVolatility
    Jun 1 at 13:49
  • $\begingroup$ Using the fact that $$ \frac{1}{\sqrt{2\pi}}\int_{\mathbb R}e^{a\omega-\frac{a^2}{2}}e^{-\frac{\omega^2}{2}}d\omega=1 $$ every such integral is very easy to solve. $\endgroup$
    – Kurt G.
    Jun 1 at 14:02
  • $\begingroup$ @ForeignVolatility the payoff is $X_T^2$. It is more a math quetion than an actual derivative $\endgroup$
    – Alejandro Andrade
    Jun 1 at 14:21
  • $\begingroup$ @KurtG. I though of that, and maybe I'm missing something, but how would you defined $a$ such that $a\omega - \frac{a^2}{2} = -\sigma^2(T-t) + 2\sigma \omega \sqrt{T-t}$ $\endgroup$
    – Alejandro Andrade
    Jun 1 at 14:31
  • $\begingroup$ 1. Pull out everything that does not contain $\omega$. 2. Define $a=2\sigma\sqrt{T-t}$. 3. Add and subtract $\frac{a^2}{2}.$ $\endgroup$
    – Kurt G.
    Jun 1 at 14:34

1 Answer 1

Reset to default
2
$\begingroup$

To long for a comment

This question should be closed as it is a basic financial question.

With the hint given in my first comment the integral \begin{align} \frac{1}{\sqrt{2\pi}}\int_\mathbb R e^{-\sigma^2(T-t) + 2\sigma\omega\sqrt{T-t} - \frac{1}{2}\omega^2} d\omega \end{align} is --setting $a=2\sigma\sqrt{T-t}$ and $\frac{a^2}{2}=2\sigma^2(T-t)$-- $$ \boxed{e^{-\sigma^2(T-t)+2\sigma^2(T-t)}\underbrace{ \frac{1}{\sqrt{2\pi}}\int_\mathbb R e^{a\omega-\frac{a^2}{2} - \frac{1}{2}\omega^2} d\omega}_{=1}=e^{\sigma^2(T-t)}.} $$

Improve this answer
$\endgroup$
1
  • $\begingroup$ I was clearly not understanding what you where saying. Thanks $\endgroup$
    – Alejandro Andrade
    Jun 1 at 16:09

Not the answer you're looking for? Browse other questions tagged or ask your own question.